Search any question & find its solution
Question:
Answered & Verified by Expert
Light is incident normally on a diffraction grating through which the first diffraction is seen at \(32^{\circ}\). In this case the second order diffraction will be
Options:
Solution:
2156 Upvotes
Verified Answer
The correct answer is:
there is no second order diffraction
Angle of first order diffraction $\left(\theta_1\right)=32^{\circ}$. We know that the angle of diffraction for the $n$th order $\left(\theta_n\right)$ is given by $d \sin \theta_n=n \lambda$.
For first order diffraction we get $d \sin 32^{\circ}=1 \times \lambda$ or, $\lambda=d \sin 32^{\circ}$.
Now for second order diffraction, $d \sin \theta_2=2 \times \lambda$ or, $d \sin \theta_2=2 \times d \sin 32^{\circ}$ or $\sin \theta_2=2 \times \sin 32^{\circ}=2 \times 0.529=1.06$. Since the sine of any angle cannot be greater than 1 , therefore there is no second order diffraction.
For first order diffraction we get $d \sin 32^{\circ}=1 \times \lambda$ or, $\lambda=d \sin 32^{\circ}$.
Now for second order diffraction, $d \sin \theta_2=2 \times \lambda$ or, $d \sin \theta_2=2 \times d \sin 32^{\circ}$ or $\sin \theta_2=2 \times \sin 32^{\circ}=2 \times 0.529=1.06$. Since the sine of any angle cannot be greater than 1 , therefore there is no second order diffraction.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.